Linear Spaces

Basis

Definition: Let $V$ be a vector space. A (finite) set of vectors $S=\{v_1, v_2, \dots, v_n\}$ is called a basis set for $V$ iff

• $Span(S)=V$
• $S$ is Linearly Independent

A (finite dimensional) linear space $V$ has many bases. All bases of a linear space have the same number of elements. This number is called the dimension of the linear space.

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Example: $V = \mathbb{R}^2$, consider the two bases:

$S_1 = \begin{Bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \end{bmatrix}\end{Bmatrix}$

1. $S_1$ is linearly independent set since $c_1\begin{bmatrix} 0 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ implies $c_1 = c_2 = 0$, $\forall c_1 , c_2 \in \mathbb{R}(F)$.
2. $Span(S_1) = V = \mathbb{R}^2$ Hence $S_1$ is a basis for $V$.

$S_2 = \begin{Bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix},\begin{bmatrix} 2 \\ 3 \end{bmatrix}\end{Bmatrix}$

1. $S_2$ is linearly independent set since $c_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ implies $c_1 = c_2 = 0$, $\forall c_1 , c_2 \in \mathbb{R}(F)$.
2. $Span(S_2) = V = \mathbb{R}^2$

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Example: $V = \mathbb{R}^2$, and $F = \mathbb{R}$, consider the base:
$B = \begin{Bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix},\begin{bmatrix} 1 \\ 1 \end{bmatrix}\end{Bmatrix}$ $y = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$ $[y]_B$=?

Solution:

1. $B$ is linearly independent set since $c_1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ implies $c_1 = c_2 = 0$, $\forall c_1 , c_2 \in \mathbb{R}(F)$.
2. $Span(B) = V = \mathbb{R}^2$
3. $[y]_B = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$
4. $y = c_1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} c_1 + c_2 \\ c_2 \end{bmatrix}$
5. $c_1 + c_2 = 2$ and $c_2 = 3$
6. $c_1 = -1$ and $c_2 = 3$
7. $[y]_B = \begin{bmatrix} -1 \\ 3 \end{bmatrix}$

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Example: $V = Span\{cos(t), sin(t)\}$, and $F = \mathbb{R}$, consider the base:
$B = \begin{Bmatrix}cos(t), sin(t)\end{Bmatrix}$ $y = cos(t-\frac{\pi}{3})$ $[y]_B$=?

Solution:

1. $B$ is linearly independent set since $c_1cos(t) + c_2sin(t) = 0$ implies $c_1 = c_2 = 0$, $\forall c_1 , c_2 \in \mathbb{R}(F)$.
2. $Span(B) = V = Span\{cos(t), sin(t)\}$
3. $[y]_B = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$
4. $y = c_1cos(t) + c_2sin(t) = cos(t-\frac{\pi}{3}) = cos(t)cos(\frac{\pi}{3}) + sin(t)sin(\frac{\pi}{3}) = \frac{1}{2}cos(t) + \frac{\sqrt{3}}{2}sin(t)$
5. $c_1 = \frac{1}{2}$ and $c_2 = \frac{\sqrt{3}}{2}$
6. $[y]_B = \begin{bmatrix} \frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{bmatrix}$

Claim: For a given basis $B$, the representation $[y]_B$ of a vector $y$ is unique.
Proof: By contradiction.
Assume $[y]_B = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$ and $[y]_B = \begin{bmatrix} d_1 \\ d_2 \end{bmatrix}$
Then $y = c_1b_1 + c_2b_2 = d_1b_1 + d_2b_2$
$c_1b_1 + c_2b_2 - d_1b_1 - d_2b_2 = 0$
$(c_1 - d_1)b_1 + (c_2 - d_2)b_2 = 0$
Since $B$ is linearly independent, $c_1 - d_1 = 0$ and $c_2 - d_2 = 0$
Hence $c_1 = d_1$ and $c_2 = d_2$ $\blacksquare$

Remark: The representation of a vector $y$ in a basis $B$ is unique. The representation of a vector $y$ in a basis $B$ is called the coordinate vector of $y$ with respect to $B$.

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Ordered Basis

Definition: Let $V$ be a vector space. An ordered set of basis vectors $S=\{v_1, v_2, \dots, v_n\}$ is called an ordered basis for $V$. If $y = (x_1,x_2,...,x_n)$ is an ordered basis for $V$, then every vector $x \in V$ can be written as a linear combination of the basis vectors as follows:

Theorem: Let $V$ be an n-dimensional vector space over $\mathbb{R}$. Let $B_1$ and $B_2$ be two bases for $V$. Then there exists a unique $n \times n$ real invertible matrix $P$ such that $[x]_{B_2} = P[x]_{B_1}$ for all $x \in V$.

Proof: By construction.

Example: Consider $V =$ polynomials of degree $\leq 2$ with coefficients in $\mathbb{R}$ and the bases:
$B_1 = \begin{Bmatrix}1, t-1, (t-1)^2\end{Bmatrix}$
$B_2 = \begin{Bmatrix}1, t, t^2\end{Bmatrix}$
Find the matrix $P$ such that $[x]_{B_1} = P[x]_{B_2}$ for all $x \in V$.

Solution:

1. $[x]_{B_1} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}$ and $[x]_{B_2} = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}$
2. $x = c_1 + c_2(t-1) + c_3(t-1)^2 = c_1 + c_2t - c_2 + c_3t^2 - 2c_3t + c_3$
3. $x = (c_1 - c_2 + c_3) + (c_2 - 2c_3)t + c_3t^2$
4. $d_1 = c_1 - c_2 + c_3$
5. $d_2 = c_2 - 2c_3$
6. $d_3 = c_3$
7. $c_1 = d_1 + d_2 + d_3$
8. $c_2 = d_2 + 2d_3$
9. $c_3 = d_3$
10. $[x]_{B_1} = \begin{bmatrix} d_1 + d_2 + d_3 \\ d_2 + 2d_3 \\ d_3 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix} = P[x]_{B_2}$
11. $P = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}$

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Example: Show that Matrix $P$ is invertible.
$~$Proof: Let $V$ be an n-dimensional vector space over $\mathbb{R^{n\times n}}$. $B_1$ and $B_2$ are bases for $V$. A vector $v$ holds,

• $[v]_{B_1} = P[v]_{B_2}$
• $\exist \ Q \in \mathbb{R^{n \times n}}$ s.t.
  • $[v]_{B_2} = Q[v]_{B_1}$
  • $[v]_{B_1} = P[v]_{B_2} = P(Q[v]_{B_1}) = (PQ)[v]_{B_1}$ implies $PQ = I$
  • similarly $QP = I$
• $P$ is invertible and $P^{-1} = Q$ $\blacksquare$

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#EE501 - Linear Systems Theory at METU